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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 751. |
Rationalise the given statements and give chemical reactions. a. Lead(II) chloride does not react with `Cl_(2)` to give `PbCl_(4)`. b. Lead(IV) chloride is highly unstable towards heat. c. Lead is known not to form an iodide, `Pbl_(4)`. |
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Answer» (a) Lead belongs to group 14 of the periodic table. The two oxidation states displayed by this group is +2 and| +4. On moving down the group, the +2 oxidation state becomes more stable and the +4 oxidation state becomes less stable. This is because of the inert pair effect. Hence, `PbCl_(4)` is much less stable than `PbCl_(2)`. However, the formation of `PbCl_(4)` takes place when chlorine gas is bubbled through a saturated solution of `PlCl_(2)`. `PbCl_(2(s))+Cl_(2(g))rarrPbCl_(4)(l)` (b) On moving down group IV, the higher oxidation state becomes unstable because of the inert pair effect. Pb(IV) is highly unstable and when heated, it reduces to Pb(II). `PbCl_(4(l))overset(Delta)rarrPbCl_(2(s))+Cl_(2(g))` (c) Lead is known not to form `PbI_(4)`. Pb (+4) is oxidising in nature and `I^(-)` is reducing in nature. A combination of Pb(IV) and iodide ion is not stable. Iodide ion is strongly reducing in nature. Pb(IV) oxidises `I^(–)" to "I^(2)` "and itself gets reduced to Pb(II)". `PbI_(4)rarrPbl_(2)+I_(2)` |
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| 752. |
Identify `(A) + N_(2) overset(Delta) rarr (B) overset(H_(2) O) rarr underset("White ppt.") ((C)darr) (D)_((g))` White precipitate `( C)` dissolves is `NaOH` solution the gas `(D)` gives white fumes with `HCl`.A. BB. AIC. GAD. C |
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Answer» Correct Answer - B `underset((A))(2AI) + N_(2) to underset((B))(2AIN)` `underset((B))(AIN) + 3H_(2)O to underset((C ))(AI(OH)_(3)) + underset((D))(NH_(3))` |
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| 753. |
Which of the noble gases can form a superfluid?A. `Xe`B. `Ne`C. `He`D. `Ar` |
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Answer» Correct Answer - 3 A superfluid is a most unusual state of matter. Normally atoms are freeto move in a gas ,can move in a more restricted way in a liquid,and can only vibrates about fixed positions in a solid. As the temperature decreases, the amount of thermal motion of atoms decrease and gases become liquids and eventually solids. At a pressure of `100 kPa` the He gas condenses at `4.2K` to form an ordinary liquid (referred to as helium `I`), but when cooled below `2.2K` the properties of the liquid (now helium II) are dramatically different for example ,helium II has zero viscocity and is an incredibly good thermal conductor ,`10^(6)` times greater than helium `I`. |
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| 754. |
Why is `H_(2)O` a liquid and `H_(2)S` a gas? |
| Answer» Hydrogen bonding occurs in `H_(2)O`. | |
| 755. |
`H_(2)S` acts only as reducing agent while `SO_(2)` can act both as a reducing agent and as an oxidising agent. How will you account for it ? |
| Answer» In `H_(2)S`, the oxidation state of sulphur is - 2. It can only increase its oxidation state but cannot decrease it. Therefore, it can act only as a reducing agent. In `SO_(2)`, the oxidation state of sulphur is + 4. It is in a position to undergo a decrease as well as an increase in the oxidation state. Thus, `SO_(2)` can act both as an oxidising agent and a reducing agent. | |
| 756. |
The compound which could not act both as oxidising and reducing agent is |
| Answer» In `SO_(2)` the oxidation state of sulphur is `+4`. Since it can undergo an increase as well as decrease in its oxidation state, it can act both as oxidising and reducing agent. On the other hand, oxidation state of sulphur in `SO_(3)` is `+6` which is the maximum it can have. Since it can only decrease its oxidation state, sulphur trioxide `(SO_(3))` can act only as an oxidising agent. | |
| 757. |
why does `NO_(2)` dimerise? |
| Answer» `NO_(2)` contains odd number of valence electron. On dimerisation it gets converted to stable `N_(2)O_(4)` with even number of electrons. | |
| 758. |
How can it be proved that `PH_(3)` is basic in nature? |
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Answer» `PH_(3)` can react with acid like Hl, `PH_(3)+Hl to PH_(4)l` As `PH_(3)` has lone pair of electron, hence it acts as lewis base. |
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| 759. |
Dioxygen gas is usually obtained in the laboratory by the thermal decomposition ofA. `KNO_(3)`B. `KMnO_(3)`C. `KCIO_(3)`D. `(NH_(4))_(2)Cr_(2)O_(7)` |
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Answer» Correct Answer - 3 The usual laboratery method consisting in heating of a mixture of potassium chlorate and maganese oxide in the ratio of `1:4KClO_(3)` evolves oxygen at `375^(@)C` but at this temperature. it melts and get converted into potassium `4KClO_(3)rarr3KClO_(4)+KCl` `KClO_(4)overset(650^(@)C)rarrKCl+2O_(2)` When `MnO_(2)` is added ,`KClO_(3)` starts giving `O_(2)` at ` 250^(@)C` thus, `MnO_(2)` acts as a catayst. `2KClO_(3)underset(heat)overset(MnO_(2))rarr2KCl+3O_(2)` `MnO_(2)` used for this purpose should be completely free from carbon as `KClO_(3)` and `C` mixture is expolsive in nature. The mixture should be first hated gently and then strongly when it is ensured that there is no vigorous reaction.`2KNO_(3)overset(Delta)rarr2KNO_(2)+O_(2)` `2KMnO_(4)overset(Delta)rarrK_(2)MnO_(4)+MnO_(2)+O_(2)` `(NH_(4))_(2)Cr_(2)O_(7)overset(Delta)rarrCr_(2)O_(3)+H_(2)O+N_(2)` `4K_(2)Cr_(2)O_(7)overset(Delta)rarr4K_(2)CrO_(4)+2Cr_(2)O_(3)+3O_(2)` |
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| 760. |
Very pure nitrogen can be obtained by :A. Thermal decomposition of `NH_(4)Cl " and " NaNO_(2)`B. Treating aqueous solution of `NH_(4)Cl " and " NaNO_(2)`C. Liquefication and fractional distillation of liquid airD. Thermal decomposition of sodium azide. |
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Answer» Correct Answer - D Very pure `N_(2)` can be obtained by the thermal decomposition of sodium azide. `underset("Sodium azide")(2NaN_(3)) overset("heat") to 2Na + 3N_(2)` |
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| 761. |
Small quantities for very pure `N_(2)`may be obtained by carefully warmingA. a mixture of `NH_(3)` and `Br_(2)`B. barium azideC. sodium azideD. both (2) and (3) |
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Answer» Correct Answer - 4 Small amounts of very pure `N_(2)(g)` can be obtained by the thermal decomposition of sodium or barium azide: `2NaN_(3)overset(300^(@)C)rarr3N_(2)(g)+2Na` `Ba(N_(3))_(2) rarr Ba+N_(2)` |
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| 762. |
Which of the following reactions is used in the preparation of `N_(2)(g)`?A. `(NH_(4))_(2)Cr_(2)O_(7)overset("heat")(rarr)`B. `NH_(3)+Ca(OCl)_(2)overset("heat")(rarr)`C. `NH_(4)Cl+NaNO_(2)overset("heat")(rarr)`D. All of these |
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Answer» Correct Answer - 4 A cylinder of `N_(2)`is th eusual source of `N_(2)`in the laboratory ,but in the laboratory ,dinitorgen is prepared by warming a concentrated solution of a ammonium nitrite made in situ from ammonium chloride and sodium nitrite until it decomposes: `NH_(4)Cl+NaNO_(3)rarrNaCl+NH_(4)NO_(2)overset("warm")rarrN_(2)+2H_(2)O` it can also be obtained by the thermal decomposition fo ammonium dischromate: `(NH_(4))_(2)Cr_(2)O_(7)overset(Delta)rarrN_(2)+4H_(2)O+Cr_(2)O_(3)` `N_(2)` is also obtained by oxidizing `NH_(3)`with calcium hypocholrite. `4NH_(3)+3Ca(OCl)_(2)rarr2N_(2)+3CaCl_(2)+6H_(2)O` |
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| 763. |
Which of the following is hydrolysed to give white precipitate of oxychloride?A. `NCl_(3)`B. `PCl_(3)`C. `BiCl_(3)`D. `AsCl_(3)` |
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Answer» Correct Answer - 3 Tricholrides readily undergo hydrolysis but the products of hydrolysis depends on the nature of the elements: `NCl_(3)+3H_(2)OrarrNH_(3)+3HOCl` `PCl_(3)+3H_(2)OrarrH_(3)PO_(3)+3HCl` `AsCl_(3)+3H_(2)OrarrH_(3)AsO_(3)+3HCl` In contrast the trichlorides of `Sb` and`Bi` are only partly and reversibly hydrolysed to give `HCl` and oxychlorides of the corresponding metal. `SbCl_(3)+H_(2)O hArr underset("oxychloride")underset("Antimony")(SbO^(+) Cl^(-))+2HCl` `BiCl_(3)+H_(2)O hArr underset(("white ppt"))underset("oxychloride")underset("Bismuth")(BiO^(+)Cl^(-))+2HCl` In accordance with`Le` Chatelier principle the addition of excess of `HCl`suuppresses the hydrolysis by shifting the equilibrium to the left. |
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| 764. |
When copper is heated with conc.`HNO_(3)` it producesA. `Cu(NO_(3))_(2)` and `N_(2)O`B. `Cu(NO_(3))_(2)` and `NO_(2)`C. `Cu(NO_(3))_(2)` and `NO`D. `Cu(NO_(3))_(2), NO` and `NO_(2)` |
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Answer» Correct Answer - 2 `Cu+4HNO_(3)("conc.")overset(Delta)rarrCu(NO_(3))_(2)+2NO_(2)+2H_(2)O` ` Cu+4HNO_(3)("dilute")overset(Delta)rarr3Cu(NO_(3))_(2)+2NO+4H_(2)O` |
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| 765. |
Which one is not a property of ozone?A. it acts as an oxidising agent in dry state.B. oxidation of `KI" into "KIO_(2).`C. PbS is oxidised to `PbSO_(4)`D. Hg is oxidised to `Hg_(2)O`. |
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Answer» Correct Answer - B Alk. KI is oxidised to potassium iodate and periodate. `KI+4O_(3)rarrunderset("pot. periodate")(KIO_(4))+4O_(2)` `KI+3O_(2)rarrunderset("pot. iodate")(KIO_(3))+3O_(2)` |
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| 766. |
The correct order of acidic strength isA. `K_(2)0 gt CaO gt MgO`B. `CO_(2) gt N_(2)O_(5) gt SO_(3)`C. `Na_(2)O gt MgO gt Al_(2)O_(3)`D. `Cl_(2)O_(7) gt SO_(2) gt P_(4)O_(10)` |
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Answer» Correct Answer - D The species `Cl_(2)O_(7), SO_(2) and P_(4)O_(10)` are the anhydrides of `HClO_(4), H_(2)SO_(3) and H_(3)PO_(4)` respectively. The acid strength of these acids follows the order `HClO_(4) gt H_(2)SO_(3) gt H_(3)PO_(4)`. The corresponding anhydrides also follow the same order. |
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| 767. |
Which of the following is the wrong statement?A. Ozone is paramagnetic gas.B. The two oxygen -oxygen bond length in ozone are identical.C. `O_(3)` molecule is bent.D. Ozone is violet-black in solid state. |
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Answer» Correct Answer - A Ozone is diamagnetic gas. |
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| 768. |
Among the following ,which one is the wrong statementA. `PH_(5)` and `BiCl_(5)` do not existB. `ppi-ppi` bonds are present in `SO_(2)`C. `SeF_(4)` and `CH_(4)` have same shapeD. `I_(3)^(+)` has bent geometery |
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Answer» Correct Answer - 3 `SeF_(4)implies` [steric number =`4`atoms+ `1` m lone pair]`implies`geometery is trigonal bipyramidal,shape is see -saw and `sp^(3) d` hybridization `CH_(4)implies` [Steric number of `4` atoms+`0` lone pair]`implies`geometery as well as shape is trtrahedral an `sp^(3)` hybridization (1) `H` is not sufficiently electronegative to contract `d`-orbitals and `Bi(+5)` is unstable due to incrt pair effect. (2) In `SO_(2)` both `ppi-ppi` and `ppi-dpi` bonds are present `I_(3)^(+) implies`[steric number=`2`atoms +`2` lone pairs ]`implies`geometery is tetrahedral shape is bent`//`angular and `sp^(3)` hybridization. |
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| 769. |
Which of the following statements is wrong ?A. Single N-N bond is stronger than the single P- P bond.B. `PH_(3)` can act as a ligand in the formation of coordination compound with transition elements.C. `NO_(2)` is paramagnetic in nature.D. Covalency of nitrogen in `N_(2)O_(5)` is four. |
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Answer» Correct Answer - A `underset(* *)overset(* *)( : N)-underset(* *) overset(* *) ( N : )` single bond is weaker than `overset(* *)underset(* *)( : P)-overset(* *)underset(* *) (P : )` bond because of greater repulsion in the lone pairs of electrons due to smaller size of nitrogen atom. |
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| 770. |
Which of the following species are isoelectronic with `N_(2)` molecule?A. `O_(2)^(+),O_(2)^(-)` and `CO^(+)`B. `CO,CN^(-)` and `NO^(+)`C. `CO,CN^(+)` and `NO^(-)`D. `CO^(+),NO` and `O_(2)^(2-)` |
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Answer» Correct Answer - 2 Isoelectronic species possess the same ekectronic configuaration because they consist of same total number of electrons. The `N_(2)` molecules `(14 e^(-)s)` is isoelectronic with `CO, CN^(-)` and `NO^(+)` |
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| 771. |
Nitrogen has unique ability to form strongA. `p_(pi)-p_(pi)` multiple bondsB. `d_(pi)-p_(pi)` multiple bondsC. `d_(pi)-d_(pi)` multiple bondsD. All of these |
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Answer» Correct Answer - 1 On account of its smaller size,`N` forms strong `P_(pi)-p_(pi)` multiple bonds but other elements of the group such as `P` As form strong `d_(pi)-p_(pi)` and `d_(pi)-d_(pi)` multiple bonds. |
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| 772. |
The maximum covalency of nitrogen isA. `6`B. `5`C. `4`D. `3` |
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Answer» Correct Answer - 3 Nitrogen is restricted to a maximum covalency of `4` because being a second period element `(n=2)`, it has only four atomic orbitals (one `2`s three `2`p) which are available for bonding . it does not have vacant `d`orbitals in its valence shell to expand its convalency beyound four. That is why it does not form pentahalides like `P` For the same reason .`N` does not function as a Lewis acid whereas `p` does so: `pF_(5) + :overset(..)underset(..)F: rarr PF_(6)^(-)` |
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| 773. |
All elements except carbon have tendency to show maximum covalency of sixA. Due to absence of vacant d-orbitalsB. Due to presence of vacant d-orbitalsC. Due to presence of partially filled d-orbitals.D. Due to presence of completely filled d-orbitals. |
| Answer» Correct Answer - B | |
| 774. |
Which of the following elements does not show allotropy ?A. NitrogenB. BismuthC. AntimonyD. Arsenic |
| Answer» Correct Answer - B | |
| 775. |
Maximum covalency of nitrogen is :A. 3B. 5C. 4D. 6 |
| Answer» Correct Answer - C | |
| 776. |
Why is `NH_(3)` a good complexing agent ? |
| Answer» `overset(* *)NH_(3)` can act as ligand due to the presence of lone pair of electrons on nitrogen atom and therfore acts as a complexing agent. | |
| 777. |
The most stable form of carbon at high temperature is X. The C - C bond length in diamond is Y while C - C bond length is graphite is Z. What are X,Y and Z respectively?A. Graphite `1.42 Å, 1.54 Å`B. Coke, `1.54 Å, 1.84 Å`C. Diamond, `1.54 Å, 1.42 Å`D. Fullerene, `1.54 Å, 1.54 Å` |
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Answer» Correct Answer - C X = diamond, `Y = 1.54 Å, Z = 1.42 Å` |
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| 778. |
Noble gases are mostly inert. Assign reasons.A. Completely filled electronic configurationB. High ionisation enthalpyC. More positive electron gain enthalpyD. All of these |
| Answer» Correct Answer - D | |
| 779. |
Noble gases were considered inert before 1962. Prior to this, Bartlett and Lohmann had previously used the highly ionization energy of `O_(2)` is `1165 " kJ mol"^(-1)`, which is almost the same as the value of `IE_(1)` for Xe `(1170 " kJ mol"^(-1))` . Experiments showed that when deep red `PtE_(6)` vapours were mixed with an equal volume of Xe, the gases combined immediately at room temperature to produce a yellow solid. Soon after this, it was found that Xe and F react directly to give Xe-fluorides. Choose the correct answer : In what molar ratio must Xe and `F_(2)` combine to give `XeF_(4)` ?A. `2 : 1`B. `1 : 2`C. `1 : 5`D. `1 : 20 ` |
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Answer» Correct Answer - C It is the correct ratio. |
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| 780. |
Liquid ammonia undergoes self-ionization ,which may be represented asA. `2NH_(3) hArr NH_(4)^(+)+NH_(2)^(+)`B. `NH_(4)OH hArr NH_(4)^(+)+OH^(-)`C. `NH_(3)+H_(2)O hArr NH_(4)^(+)+OH^(-)`D. `NH_(3)+H_(2)O hArr NH_(4)^(+) +OH^(-)` |
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Answer» Correct Answer - 4 Liquids ammonia is ionized to a slight extent: `2NH_(3) hArr NH_(4)^(+)+NH_(2)^(-)` of. `2H_(2)O hArr H_(3)O^(+)+OH^(-)` Whereas in water acids are substances that provide hydroxoide ions and alkalis are substances that provide hydroxylions, ions in liquid ammonia acids give to ammonium ions, (e.g. ammonium salst) and give rise to amide ions, (e.g. ammonium salts) and bases give rise to amide ions (e.g. ionic amides). Thus in liuid ammonia an ammono-acid neutralises an ammono-base to goce a salt and ammonia (an extact parallel with aqueous systems).e.g. `underset("ammono -acid")(NH_(4)^(+)Cl)+underset("ammono- base")(Na^(+)NH_(2)^(-))rarr underset("salt")(Na^(+)Cl^(-))+underset("ammonia")(2NH_(3))` or `NH_(4)^(+)+NH_(2)^(-)rarr2NH_(3)` Ionic nitrides such as lithium nitride function as bases in liquid ammonia: `3NH_(4)^(+)Cl^(-)+(Li^(+))_(3)N^(3-)rarr3Li^(+)Cl^(-)+4NH_(3)` or `3NH_(4)^(+)+N^(3-)rarr4NH_(3)` More parallels with aqueous systems exist, for instance hydrolysis and ammonolysis are essentially very similar processes `SiCl_(4)overset("hydrolysis")rarrSi(OH)_(4)overset("heat")rarrSiO_(2)+2H_(2)O` `3SiCl_(4)overset("ammonolysis")rarr 3Si(NH_(2))_(4)overset("heat")rarr Si_(3)N_(4)+8NH_(3)` |
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| 781. |
Name two elements of carbon family which have nearly same catenation tendency. |
| Answer» Correct Answer - Ge and Sn. | |
| 782. |
Name the element in carbon family that can be afected by water. |
| Answer» Carbon, silicon and germanium. | |
| 783. |
`Cu+4HNO_(3)("Conc.") to` |
| Answer» `Cu+4HNO_(3)("conc.") overset("heat") to CuSO_(4) +SO_(2)+2H_(2)O` | |
| 784. |
Which element of the carbon family does not dissolve in aqueous NaOH to evolve hydrogen gas? |
| Answer» Correct Answer - Carbon. | |
| 785. |
`Cu+H_(2)SO_(4)("Conc.") to ` |
| Answer» `Cu+2H_(2)SO_(4)("conc.") overset("heat") to CuSO_(4)+SO_(2)+2H_(2)O` | |
| 786. |
`(NH_(4))_(2)Cr_(2)O_(7) overset("heat") to` |
| Answer» `(NH_(4))_(2)Cr_(2)O_(7) overset("heat")to Cr_(2)O_(3)+N_(2)+4H_(2)O` | |
| 787. |
The incorrect statement among the following isA. Reducing character of hydrides of group 15 increases down the groupB. Basicity of hydrides of group 15 increases down the groupC. Phosphorus and arsenic can form `p pi- d pi` bondD. `NCl_(5)` does not exist |
| Answer» Correct Answer - B | |
| 788. |
In the reaction `Cu+ underset((Conc.))(H_(2)SO_(4))to CuSO_(4) +SO_(2) +H_(2)O` `H_(2)SO_(4)` behaves w.r.t. Cu asA. Dehydrating agentB. oxidising agentC. An acidD. All of these |
| Answer» Correct Answer - B | |
| 789. |
`P_(4)+HNO_(3)("Conc.") to` |
| Answer» `P_(4)+20HNO_(3) overset("Conc.")to 4H_(3)PO_(4) +20NO_(2)+4H_(2)O` | |
| 790. |
`Cu+underset((conc.))(HNO_(3))to`products the products formed in above reaction isA. `CuO+NO`B. `Cu(NO_(3))_(2)+N_(2)O`C. `Cu(NO_(3))_(2)+NO_(2)`D. All of these can form |
| Answer» Correct Answer - C | |
| 791. |
`Ca_(3)(PO_(4))_(2) +H_(2)SO_(4) to` |
| Answer» `Ca_(3)(PO_(4))_(2)+3H_(2)SO_(4) to 3 CaSO_(4)+2H_(3)PO_(4)` | |
| 792. |
`HNO_(3) overset(P_(4)O_(10)) underset("Heat") to` |
| Answer» `2HNO_(3) underset("Heat")overset(P_(4)O_(10)) to N_(2)O_(5)+H_(2)O` | |
| 793. |
`H_(4)P_(2)O_(7) overset("Heat")to` |
| Answer» `underset("Pyrophosphoric acid")(H_(4)P_(2)O_(7)) overset("Heat")to underset("Metaphosphoric acid")(2HPO_(3))+H_(2)O` | |
| 794. |
`(A)+SbF_(5)to[XeF_(3)]^(+)[SbF_(6)]^(-)` Compound (A) isA. `XeF_(2)`B. `XeF_(4)`C. `XeF_(6)`D. Both 1 and 2 |
| Answer» Correct Answer - B | |
| 795. |
When chlorine water is added to an aqueous solution of sodium halide in the presence of chloroform, violet colouration is obtained. When more of chlorine water is added, the violet colour disappears and the solution becomes colourless. This confrims that sodium halide is :A. ChlorideB. fluorideC. BromideD. Iodide. |
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Answer» Correct Answer - D Sodium halide is Nal. Chlorine displaces iodine form Nal which imparts violet colour - ation to chloroform layer. However, excess of chlorine water dissolves iodine to form a colourless solution. `2Na+Cl_(2) to 2NaCl+l_(2)` `l_(2)+5Cl_(2)+6H_(2)O to underset(("Colourless"))(2HlO_(3))+10HCl` |
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| 796. |
`XeF_(4) +SbF_(5) to ` |
| Answer» `XeF_(4)+SbF_(5) to [XeF_(3)]^(+)[SbF_(6)]^(-)` | |
| 797. |
`SCl_(4)+AgF to` |
| Answer» `SCl_(4)+4AgF to 4AgCl+SF_(4)` | |
| 798. |
`HPO_(3) overset("Heat") to` |
| Answer» `2HOP_(3) overset("Heat")to P_(2)O_(5)+H_(2)O` | |
| 799. |
Componds `(A)` and `B` are treated with dilute `HCl` separately. The gases liberated are `Y` and `Z` respectively. `Y` turns acidified `K_2 Cr_2 O_7` paper green while `Z` turns lead acetate paper black. The compounds `A` and `B` are respectively :A. `Na_(2)S " and " Na_(2)SO_(3)`B. `Na_(2)SO_(3) " and " Na_(2)S`C. `NaCl " and" Na_(2)CO_(3)`D. `Na_(2)SO_(3) " and " na_(2)SO_(4)` |
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Answer» Correct Answer - B `underset((A))Na_(2)SO_(3)+2HCl (aq) to 2NaCl+H_(2)O+underset((X))SO_(2)` `K_(2)Cr_(2)O_(7)+H_(2)SO_(4) + underset((X))3SO_(2) to underset(("Green"))(Cr_(2)(SO_(4))_(3))+K_(2)SO_(4)+H_(2)O` `underset((B))(Na_(2)S)+HCl(aq) to 2NaCl+underset((Y))(H_(2)S)` `underset((Y))(H_(2)S)+Pb(CH_(3)COO)_(2) to underset(("Black"))(PbS+2CH_(3))COOH` |
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| 800. |
Hybridisation and shape of `BrF_(5)` is : |
| Answer» Correct Answer - `sp^(3)d^(2)` | |